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balandron [24]
3 years ago
10

Calculate the concentration (in mol/l) of 33% by weight (33 g naoh per 100 g of solution) naoh solution. (the density of the 33%

by weight solution is 1.35 g/ml).
Chemistry
1 answer:
Shtirlitz [24]3 years ago
7 0
Molarity is measured in moles per Liter. If there are 1.35 g/mL, find out how many grams there are in a liter of solution.

If there are 1000 mL in one liter, we can multiply by 1000 to get g/L

1.35 g/mL x 1 Liter/1000 mL = 1350 g per Liter of solution

By weight, the NaOH is 33% or .33

1350 g x .33 = 445.5 g of NaOH

Molar mass of NaOH is 39.997 g

445.5 g  x 1 mol NaOH/39.997 g = 11.13833538 moles per Liter

Rounded to significant figures, the answer is 11 mol/L NaOH


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Tim wants to store 0.122 moles of helium gas at 203 kPa and 401 K. Which is the volume of the helium gas needed for storage?
algol13

 The  volume  of Helium gas needed for storage  is 2.00 L (answer C)


 <u><em> calculation</em></u>

 The volume of Helium is calculated using ideal gas equation

That is Pv =nRT

where;

 P( pressure) = 203 KPa

V(volume)=?

n(number  of moles) =  0.122 moles

R(gas constant) = 8.314  L.Kpa/mol.K

T(temperature)= 401 K

make V the  subject of the formula  by diving both side  by P

V=nRT/p

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3 0
3 years ago
In the reaction mg(s)+2hcl(aq) h2(g)+mgcl2(aq), how many grams of hydrogen gas will be produced from 125.0 ml of a 6.0m hcl in a
Kruka [31]
6.0m(mol/kg) of HCl
125mL H2O = 0.125kg
6mol/kg = n mol/0.125kg, n = 0.75mol
When 0.75mol of HCl reacts, 0.75/2=0.375mol of H2 is produced. H2 = 2g/mol
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6 0
3 years ago
Read 2 more answers
Consider the process used to produce iron metal from its ore.
Dmitry_Shevchenko [17]

Answer:

223.4 grams of iron can be produced from 2.5 moles of Fe2O3 and 6.0 moles of CO.

Explanation:

The balanced reaction is:

Fe₂O₃ (s) + 3 CO(g) → 2 Fe(s) + 3 CO₂ (g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • Fe₂O₃: 1 mole
  • CO: 3 moles
  • Fe: 2 moles
  • CO₂: 3 moles

Being:

  • Fe: 55.85 g/mole
  • O: 16 g/mole
  • C: 12 g/mole

the molar mass of the compounds participating in the reaction is:

  • Fe₂O₃: 2*55.85 g/mole + 3*16 g/mole= 159.7 g/mole
  • CO: 12 g/mole + 16 g/mole= 28 g/mole
  • Fe: 55.85 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole

Then, by stoichiometry of the reaction, the following quantities participate in the reaction:

  • Fe₂O₃: 1 mole* 159.7 g/mole= 159.7 g
  • CO: 3 moles* 28 g/mole= 84 g
  • Fe: 2 moles* 55.85 g/mole= 111.7 g
  • CO₂: 3 moles* 44 g/mole= 132 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

So, first of all, you can apply the following rule of three: if by reaction stoichiometry 1 mole of Fe₂O₃ reacts with 3 moles of CO, then 2.5 moles of Fe₂O₃ react with how many moles of CO?

moles of CO=\frac{2.5 moles of Fe_{2} O_{3}*3 moles of CO }{1 mole of Fe_{2} O_{3}}

moles of CO= 7.5

But 7.5 moles of CO are not available, 6.0 moles are available. Since you have less moles than you need to react with 2.5 moles of Fe₂O₃, CO will be the limiting reagent.

Now you can apply the following rule of three: if by reaction stoichiometry 3 moles of CO produce with 111.7 grams of Fe, then 6 moles of CO will produce how much mass of Fe?

mass of Fe=\frac{6 moles of CO*111.7 grams of Fe}{3 moles of CO}

mass of Fe= 223.4 grams

<u><em>223.4 grams of iron can be produced from 2.5 moles of Fe2O3 and 6.0 moles of CO.</em></u>

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Which equation shows how to solve for enthalpy of solution based on the information in tables?​
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Answer: B. triH sol Mgl2= -triHlat+ triHhydr Mg^2+ 2triHhydr^l-

Explanation:

Just did it and it was right

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